POJ-1328（贪心）

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 
 
Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 

The input is terminated by a line containing pair of zeros 

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

/*题解：贪心策略以小岛为圆心画圆，与X轴相交两点再按照左端点从小到大排序最后贪心过一遍就OK了*/#include <iostream>  #include<math.h>#include<algorithm>#include<stdio.h>using namespace std;struct pos{int x, y;}nowpos[1005];int flag = 1;struct node{double left, right;}nownode[1005];int d, n;bool cmp(node ss, node tt){return ss.left < tt.left;}void Cal(){for (int i = 0; i < n; i++){nownode[i].left = nowpos[i].x - ((double)sqrt(pow(d,2) - pow(nowpos[i].y,2)));nownode[i].right = nowpos[i].x + ((double)sqrt(pow(d, 2) - pow(nowpos[i].y, 2)));}}int FinaAns(){int cnt = 1;sort(nownode, nownode + n, cmp);//按照左端点从小到大排序double temp = nownode[0].right;//贪心算法for (int i = 1; i < n; i++){//没注意到就Wrong了，重点^_^||| //后面的右端点可以直接小于之前的右端点//这样就要更新右端点if (nownode[i].right < temp){temp = nownode[i].right;}else if (temp < nownode[i].left){cnt++;temp = nownode[i].right;}}return cnt;}int main(){int tempcount = 0;while (cin >> n >> d){flag = 1;if (n == 0 && d == 0)break;for (int i = 0; i < n; i++){scanf("%d%d", &nowpos[i].x, &nowpos[i].y);if (nowpos[i].y > d){flag = 0;}}if (flag == 0){//找了半天的错原来就是这里少了个Case，Q_Q printf("Case %d: -1\n", ++tempcount);}else{Cal();//计算printf("Case %d: %d\n", ++tempcount, FinaAns());//输出答案}}return 0;}